I was trying to understand the proof for the following proposition.

Proposition: If $\{f_n\}$ is a sequence of $\bar{\mathbb{R}}$ valued measurable functions on $(X,\mathcal{M})$, then the functions

$$g_1(x) = \sup_j f_j(x)$$ $$g_2(x) = \inf_j f_j(x)$$ $$g_3(x) = \lim_{j\rightarrow \infty} \sup f_j(x)$$ $$g_4(x) = \lim_{j\rightarrow \infty} \inf f_j(x)$$

are all measurable.

In the proof, it is given that $g_1^{-1}((a,\infty]) = \bigcup_1^{\infty}f_j^{-1}((a,\infty])$ and $g_2^{-1}([\infty,a)) = \bigcup_1^{\infty}f_j^{-1}([\infty,a)))$.

I am unable to understand these inverses of $g$. Can you please explain this?

  • 1
    $\begingroup$ It's probably not the inverse of $g$, but rather the preimage; that is, $g_1^{-1}((a, \infty]) = \{x: g_1(x) \in (a, \infty]\}$ $\endgroup$ – pjs36 Mar 22 '15 at 14:18
  • $\begingroup$ oh ok . But how does it equal to the union of preimages of f ? A bit confused ... $\endgroup$ – Rusty Mar 22 '15 at 14:24
  • 2
    $\begingroup$ Notice my edits: I change {$f_n$} to $\{f_n\}$ (with the curly braces inside of MathJax) and $Sup_j f_j(x)$ to $\displaystyle\sup_j f_j(x)$ and a few other similar things. ${}\qquad{}$ $\endgroup$ – Michael Hardy Mar 22 '15 at 14:24

Recall that $g_1^{-1}((a,\infty])$ means the set $\{x\in X:g_1(x)\in (a,\infty]\}$ (called the preimage of the set $(a,\infty]$ under $g_1$).

We are told that $g_1(x)=\sup\{f_j(x)\}$. This tells us two things:

  • for all $x\in X$, and for all $j$, $f_j(x)\le g_1(x)$.
  • if $y$ is such that for all $x\in X$ and all $j$, $f_j(x)\le y$, then $g_1(x)\le y$

(This is just the definition of a least upper bound.)

Now in order to show that $$ g_1^{-1}((a,\infty])=\bigcup f_j^{-1}((a,\infty]) $$ we need to show that:

  • $\bigcup f_j^{-1}((a,\infty]) \subset g_1^{-1}((a,\infty])$ - equivalently, for each $j$, $f_j^{-1}((a,\infty])\subset g_1^{-1}((a,\infty])$.
  • $g_1^{-1}((a,\infty]) \subset \bigcup f_j^{-1}((a,\infty])$ - equivalently, for each $x$ such that $g_1(x)\in (a,\infty]$, there exists some $j$ such that $f_j(x)\in (a,\infty]$.

I think you can prove both those things yourself. Then try and use the same argument for $g_2^{-1}([-\infty,a))$.


To say that $$ g_1^{-1}((a,\infty]) = \bigcup_1^\infty f_j^{-1}((a,\infty]) $$ is to say that for every $x$ in the domain, $$ g_1(x)>a\text{ if and only if there is at least one $j$ for which }f_j(x)>a. $$ i.e. $$ \sup_j f_j(x)>a\text{ if and only if there is at least one $j$ for which }f_j(x)>a. $$ i.e. the smallest upper bound of a set is $>a$ if and only if at least one member of the set is $>a$. Consider both "if" and "only if" in the light of the definition of smallest upper bound.


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